Problem: Among all pairs of real numbers $(x, y)$ such that $\sin \sin x = \sin \sin y$ with $-10 \pi \le x, y \le 10 \pi$, Oleg randomly selected a pair $(X, Y)$.  Compute the probability that $X = Y$.
Solution: The function $\sin x$ is increasing on the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right],$ so it is increasing on the interval $[-1,1].$  Hence,
\[\sin \sin x = \sin \sin y\]implies $\sin x = \sin y.$  In turn, $\sin x = \sin y$ is equivalent to $y = x + 2k \pi$ or $y = (2k + 1) \pi - x$ for some integer $k.$  Note that for a fixed integer $k,$ the equations $y = x + 2k \pi$ and $y = (2k + 1) \pi - x$ correspond to a line.  These lines are graphed below, in the region $-10 \pi \le x,$ $y \le 10 \pi.$

[asy]
unitsize(0.15 cm);

pair A, B, C, D;
int n;

A = (-10*pi,10*pi);
B = (10*pi,10*pi);
C = (10*pi,-10*pi);
D = (-10*pi,-10*pi);

draw(B--D,red);

for (n = 1; n <= 9; ++n) {
  draw(interp(A,D,n/10)--interp(A,B,n/10),red);
	draw(interp(C,D,n/10)--interp(C,B,n/10),red);
}

for (n = 1; n <= 19; ++n) {
  if (n % 2 == 1) {
	  draw(interp(D,C,n/20)--interp(D,A,n/20),blue);
		draw(interp(B,C,n/20)--interp(B,A,n/20),blue);
	}
}

draw(A--B--C--D--cycle);
[/asy]

There are 200 points of intersection.  To see this, draw the lines of the form $x = n \pi$ and $y = n \pi,$ where $n$ is an integer.

[asy]
unitsize(0.15 cm);

pair A, B, C, D;
int n;

A = (-10*pi,10*pi);
B = (10*pi,10*pi);
C = (10*pi,-10*pi);
D = (-10*pi,-10*pi);

draw(B--D,red);

for (n = 1; n <= 9; ++n) {
  draw(interp(A,D,n/10)--interp(A,B,n/10),red);
	draw(interp(C,D,n/10)--interp(C,B,n/10),red);
}

for (n = 1; n <= 19; ++n) {
  if (n % 2 == 1) {
	  draw(interp(D,C,n/20)--interp(D,A,n/20),blue);
		draw(interp(B,C,n/20)--interp(B,A,n/20),blue);
	}
}

for (n = -9; n <= 9; ++n) {
  draw((-10*pi,n*pi)--(10*pi,n*pi),gray(0.7));
	draw((n*pi,-10*pi)--(n*pi,10*pi),gray(0.7));
}

draw(A--B--C--D--cycle);
[/asy]

These lines divide the square into 400 smaller squares, exactly half of which contain an intersection point.  Furthermore, exactly 20 of them lie on the line $y = x,$ so the probability that $X = Y$ is $\frac{20}{400} = \boxed{\frac{1}{20}}.$